Problem: Solve for $r$, $ -\dfrac{r - 6}{15r^2} = -\dfrac{2}{3r^2} + \dfrac{3}{3r^2} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15r^2$ $3r^2$ and $3r^2$ The common denominator is $15r^2$ The denominator of the first term is already $15r^2$ , so we don't need to change it. To get $15r^2$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{2}{3r^2} \times \dfrac{5}{5} = -\dfrac{10}{15r^2} $ To get $15r^2$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{3}{3r^2} \times \dfrac{5}{5} = \dfrac{15}{15r^2} $ This give us: $ -\dfrac{r - 6}{15r^2} = -\dfrac{10}{15r^2} + \dfrac{15}{15r^2} $ If we multiply both sides of the equation by $15r^2$ , we get: $ -r + 6 = -10 + 15$ $ -r + 6 = 5$ $ -r = -1 $ $ r = 1$